Properties of Triangles: Very Difficult Problems with Solutions

Solution:
Let [tex]CA=CB[/tex] and let the vertex angle be [tex]\gamma[/tex]; the base angles are [tex]90^\circ-\frac{\gamma}{2}[/tex]. Let [tex]AH[/tex] be the altitude to line [tex]BC[/tex]. In the right triangle [tex]ABH[/tex] the angle with the base is [tex]\angle BAH=90^\circ-\angle B=\frac{\gamma}{2}[/tex].
If [tex]\gamma\lt 90^\circ[/tex], [tex]H[/tex] lies on side [tex]BC[/tex] and [tex]\angle CAH=90^\circ-\gamma[/tex]:
[tex]90^\circ-\gamma=\dfrac{\gamma}{2}-12^\circ \Rightarrow \gamma=68^\circ[/tex]
If [tex]\gamma\gt 90^\circ[/tex], [tex]H[/tex] lies on the extension of [tex]BC[/tex] beyond [tex]C[/tex] and [tex]\angle CAH=\gamma-90^\circ[/tex]:
[tex]\gamma-90^\circ=\dfrac{\gamma}{2}-12^\circ \Rightarrow \gamma=156^\circ[/tex]
(For [tex]\gamma=90^\circ[/tex] the altitude is the leg itself.) So the vertex angle is [tex]68^\circ[/tex] or [tex]156^\circ[/tex].

Solution:
Let [tex]\angle A=\alpha[/tex], [tex]\angle B=\beta[/tex], [tex]\angle C=\gamma[/tex]. By the exterior angle theorem:
[tex]\angle AMC=\beta+\dfrac{\alpha}{2}, \angle BPC=\alpha+\dfrac{\beta}{2}[/tex]
A line forms angles of [tex]75^\circ[/tex] and [tex]105^\circ[/tex] with a side, so each of these angles is [tex]75^\circ[/tex] or [tex]105^\circ[/tex]. Both equal to [tex]75^\circ[/tex]: subtracting gives [tex]\alpha=\beta[/tex], so [tex]\alpha=\beta=50^\circ[/tex], [tex]\gamma=80^\circ[/tex]. Both equal to [tex]105^\circ[/tex]: [tex]\alpha=\beta=70^\circ[/tex], [tex]\gamma=40^\circ[/tex].
If [tex]\beta+\frac{\alpha}{2}=75^\circ[/tex] and [tex]\alpha+\frac{\beta}{2}=105^\circ[/tex], adding gives [tex]\alpha+\beta=120^\circ[/tex] and subtracting gives [tex]\alpha-\beta=60^\circ[/tex], so [tex]\alpha=90^\circ[/tex], [tex]\beta=30^\circ[/tex], [tex]\gamma=60^\circ[/tex]. The opposite case gives [tex]\alpha=30^\circ[/tex], [tex]\beta=90^\circ[/tex], [tex]\gamma=60^\circ[/tex].

Solution:
Let [tex]\angle CBD=x[/tex] and [tex]\angle BAC=2x[/tex]. Then:
[tex]\angle ADB=140^\circ-2x, \angle C=140^\circ-3x[/tex]
In triangle [tex]ABD[/tex], [tex]BD[/tex] lies opposite the angle [tex]2x[/tex] and [tex]AB[/tex] opposite [tex]140^\circ-2x[/tex]. In triangle [tex]BDC[/tex], [tex]BD[/tex] lies opposite [tex]140^\circ-3x[/tex] and [tex]DC[/tex] opposite [tex]x[/tex]. If [tex]x\gt 35^\circ[/tex], then [tex]2x\gt 140^\circ-2x[/tex], so [tex]BD\gt AB[/tex], and [tex]140^\circ-3x\lt x[/tex], so [tex]BD\lt DC=AB[/tex], a contradiction. If [tex]x\lt 35^\circ[/tex], both inequalities reverse, again a contradiction. Hence:
[tex]x=35^\circ[/tex]
Check: the angles of triangle [tex]ABC[/tex] are [tex]70^\circ[/tex], [tex]75^\circ[/tex] and [tex]35^\circ[/tex], and [tex]AB=BD=DC[/tex].

Solution:
Let [tex]CA=CB[/tex], the base angles be [tex]\alpha[/tex] and the vertex angle [tex]\gamma=180^\circ-2\alpha[/tex]. The line through [tex]A[/tex] meets [tex]BC[/tex] at [tex]D[/tex]; let [tex]\angle BAD=x[/tex], [tex]0\lt x\lt \alpha[/tex]. Triangle [tex]ABD[/tex] has angles [tex]x, \alpha, 180^\circ-\alpha-x[/tex], and triangle [tex]ADC[/tex] has angles [tex]\alpha-x, \gamma, \alpha+x[/tex].
Triangle [tex]ABD[/tex] is isosceles when [tex]180^\circ-\alpha-x=\alpha[/tex], i.e. [tex]x=\gamma[/tex], or when [tex]x=180^\circ-\alpha-x[/tex], i.e. [tex]x=90^\circ-\frac{\alpha}{2}[/tex]. Triangle [tex]ADC[/tex] is isosceles when [tex]\alpha-x=\gamma[/tex] or [tex]\alpha+x=\gamma[/tex].
[tex]x=\gamma[/tex] and [tex]\alpha-x=\gamma[/tex]: [tex]\alpha=2\gamma[/tex], so [tex]\gamma=36^\circ[/tex], [tex]\alpha=72^\circ[/tex]. [tex]x=\gamma[/tex] and [tex]\alpha+x=\gamma[/tex]: [tex]\alpha=0[/tex], impossible.
[tex]x=90^\circ-\frac{\alpha}{2}[/tex] and [tex]\alpha-x=\gamma[/tex]: [tex]\frac{3\alpha}{2}-90^\circ=180^\circ-2\alpha[/tex], so [tex]\alpha=\frac{540^\circ}{7}[/tex], [tex]\gamma=\frac{180^\circ}{7}[/tex] (and [tex]x=\frac{360^\circ}{7}\lt \alpha[/tex]). [tex]x=90^\circ-\frac{\alpha}{2}[/tex] and [tex]\alpha+x=\gamma[/tex]: [tex]\alpha=36^\circ[/tex], but then [tex]x=72^\circ\gt \alpha[/tex], impossible.

Solution:
Triangle [tex]ADC[/tex] is right-angled with a [tex]45^\circ[/tex] angle, so [tex]AD=CD[/tex]. Since [tex]D[/tex] lies on the bisector of [tex]\angle AMB[/tex], it is equidistant from [tex]MA[/tex] and [tex]MB[/tex]: if [tex]DP\perp AM[/tex] and [tex]DQ\perp BC[/tex], then [tex]DP=DQ[/tex].
The right triangles [tex]ADP[/tex] and [tex]CDQ[/tex] have equal hypotenuses ([tex]AD=CD[/tex]) and equal legs ([tex]DP=DQ[/tex]), so they are congruent and [tex]\angle DAP=\angle DCQ[/tex]:
[tex]\angle BAM=\angle DCB=90^\circ-\angle B[/tex]
[tex]\angle AMB=180^\circ-\angle B-(90^\circ-\angle B)=90^\circ[/tex]
So [tex]AM[/tex] is the altitude from [tex]A[/tex].

Solution:
[tex]\angle BAD=60^\circ-45^\circ=15^\circ[/tex] (exterior angle of triangle [tex]ABD[/tex]). Drop [tex]CK\perp AD[/tex]. In the right triangle [tex]DKC[/tex], [tex]\angle DCK=30^\circ[/tex], so [tex]DK=\frac{DC}{2}=BD[/tex]. Since [tex]AD\gt BD[/tex] (in triangle [tex]ABD[/tex], [tex]45^\circ\gt 15^\circ[/tex]), [tex]K[/tex] lies on segment [tex]AD[/tex].
Triangle [tex]BDK[/tex] is isosceles with [tex]\angle BDK=120^\circ[/tex], so [tex]\angle DBK=\angle DKB=30^\circ[/tex]. Then [tex]\angle KBC=30^\circ=\angle KCB[/tex], so [tex]KB=KC[/tex]. Also [tex]\angle ABK=45^\circ-30^\circ=15^\circ=\angle BAK[/tex], so [tex]KA=KB[/tex]. Hence [tex]KA=KC[/tex], the right triangle [tex]AKC[/tex] is isosceles and [tex]\angle KCA=45^\circ[/tex]:
[tex]\angle ACB=30^\circ+45^\circ=75^\circ[/tex]

Solution:
[tex]D[/tex] lies on the bisector of [tex]\angle B[/tex], so it is equidistant from lines [tex]BA[/tex] and [tex]BC[/tex]. [tex]D[/tex] lies on the bisector of [tex]\angle AKC[/tex], so it is equidistant from lines [tex]KA[/tex] and [tex]KC[/tex], and line [tex]KC[/tex] is line [tex]BC[/tex]. Hence [tex]D[/tex] is equidistant from lines [tex]AB[/tex] and [tex]AK[/tex], so line [tex]AD[/tex], i.e. [tex]AC[/tex], bisects one of the angles formed by these two lines.
[tex]AC[/tex] lies outside [tex]\angle BAK[/tex], so it bisects the angle between [tex]AK[/tex] and the extension [tex]AB'[/tex] of [tex]BA[/tex] beyond [tex]A[/tex]: [tex]\angle KAC=\angle CAB'[/tex]:
[tex]\dfrac{\angle BAC}{2}=180^\circ-\angle BAC \Rightarrow \angle BAC=120^\circ[/tex]

Solution:
The base angles are [tex]50^\circ[/tex]. Let [tex]H[/tex] be the midpoint of [tex]AB[/tex]; [tex]CH[/tex] is the axis of symmetry of the triangle. Let line [tex]BM[/tex] meet [tex]CH[/tex] at [tex]N[/tex] ([tex]M[/tex] lies between [tex]B[/tex] and [tex]N[/tex]). Since [tex]NA=NB[/tex], [tex]\angle NAB=30^\circ[/tex], so [tex]\angle NAM=20^\circ[/tex] and [tex]\angle NAC=50^\circ-30^\circ=20^\circ[/tex].
[tex]\angle ANB=120^\circ[/tex] and [tex]\angle ANH=60^\circ[/tex], so [tex]\angle ANC=120^\circ=\angle ANM[/tex]. Triangles [tex]ANC[/tex] and [tex]ANM[/tex] have the common side [tex]AN[/tex] and equal adjacent angles, so they are congruent and [tex]AM=AC[/tex]. In the isosceles triangle [tex]ACM[/tex], [tex]\angle CAM=50^\circ-10^\circ=40^\circ[/tex]:
[tex]\angle AMC=\dfrac{180^\circ-40^\circ}{2}=70^\circ[/tex]

Solution:
Let [tex]\angle A=\alpha[/tex] and [tex]\angle B=\beta[/tex]. Since [tex]\alpha\lt \beta[/tex], the points on [tex]AB[/tex] are in the order [tex]A[/tex], [tex]M[/tex], [tex]H[/tex], [tex]B[/tex]. Take [tex]D[/tex] on ray [tex]HB[/tex] beyond [tex]B[/tex] with [tex]BD=BC[/tex]. Then [tex]HD=HB+BC=MH[/tex], so [tex]CH[/tex] is the perpendicular bisector of [tex]MD[/tex] and [tex]CD=CM[/tex].
Triangle [tex]BCD[/tex] is isosceles and its exterior angle at [tex]B[/tex] is [tex]\angle ABC=\beta[/tex], so [tex]\angle BDC=\frac{\beta}{2}[/tex]. Since [tex]CM=CD[/tex], [tex]\angle CMB=\angle CDM=\frac{\beta}{2}[/tex]. On the other hand, [tex]MA=MC[/tex] (median to the hypotenuse), so the exterior angle [tex]CMB[/tex] of triangle [tex]AMC[/tex] equals [tex]2\alpha[/tex]:
[tex]2\alpha=\dfrac{\beta}{2} \Rightarrow \beta=4\alpha, \alpha+4\alpha=90^\circ \Rightarrow \alpha=18^\circ[/tex]