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Congruent Triangles Problems by Side-Angle-Side

Problem 1
Triangle ABC – isosceles, CD is bisector to the base AB. Prove that ACD is congruent to BCD
Solution:
Prove that the two triangles are congruent by Side-Angle-Side. From the condition we have angle ACD = DCB (CD is bisector); AC = BC (triangle ABC isosceles); CD is in both triangles. So triangles ACD and BCD have congruent two sides and the included angle. Therefore triangle ACD and BCD are congruent.

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Problem 2
Prove that the altitude, the bisector and the median to the base of a isosceles triangle match.
Solution:
From the previous problem we have the CD is bisector in the isosceles triangle ABC and we proved that triangles ACD and BCD are congruent. From the congruence follows that angle ADC = CDB but they are contiguous therefore their sum is 180°, from where angle ADC = CDB = 90°, which show that CD is an altitude. From the congruence of the two triangle we have that AD = BD, i.e. CD is a median.

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Problem 3
Prove that the angles in the base of every isosceles triangle are equal.
Solution:
We use again problem 1 and that triangles ACD and BCD are congruent. The angles in the base of isosceles triangle are equal because they are corresponding angles in congruent triangles.

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Problem 4
Calculate the perimeter of isosceles triangle ABC if the perimeter of >ADC is 18 cm., and CD = 6 cm. and AD = BD (fig.5)
Solution:
The perimeter of triangle ADC = AC + CD + AD = 18 <=> AC + 6 + AD = 18 <=> AC + AD = 12
Because AC = BC (the triangle is isosceles) and AD = DB follows that AC + AD = DB +BC = 12
The perimeter of triangle ABC = AB + AC + BC = AD + DB + AC + BC = 12 + 12 = 24cm.

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Problem 5
Prove that a straight line which cuts equal segments from the shoulders of an angle is perpendicular to the bisector of that angle.
Solution:

Lets take that the straight line cuts from the shoulders of angle AOB equal segments OC = OD. Then triangle OCD is isosceles and OF is bisector to the base. According to problem 2 OF is an altitude, i.e. OF is perpendicular to α

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Problem 6
Prove that if the diagonals of a quadrangle halve each other every two opposite sides are equal.
Solution:

For the quadrangle ABCD we know that AO = OC è BO = OD. Then triangles AOD and BOC are congruent (side-angle-side, AO = OC; BO = OD and angles DOA = BOC – apex) therefore AD = BC Analogically triangles AOB and DOC are congruent from where AB = CD.

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Problem 7
Prove that if triangle ABC is congruent to A₁B₁C₁ and for points M and M₁ from sides AB and A₁B₁ it is in force that AM = A₁M₁ CM = C₁M₁ and angle BMC = B₁M₁C₁
Solution:

Because triangle ABC is congruent to triangle A₁B₁C₁ AB = A₁B₁, BC = B₁C₁, CA = C₁A₁ and angles A = A₁, B = B₁, C = C₁. Then triangle AMC is congruent to triangle A₁M₁C₁ (side-angle-side), from where CM = C₁M₁. MB = M₁B₁ because from the equal segments AB and A₁B₁ we subtract the equal segments AM and A₁M₁. Angles BMC and B₁M₁C₁ are equal because they are equal contiguous angles to angles AMC and A₁M₁C₁, which are equal (triangle AMC is congruent to A₁M₁C₁)

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Problem 8
We have triangle ABC (AC < BC). Outwardly are built:
squares BMNC and CPQA. Prove that AN = BP

Solution:
For triangles BPC and ACN BC = CN (BMNC – square);
PC = AC (CPQA - square) angle BCP = ACN = 90° + ACB. Therefore triangles are congruent from where AN = BP

Congruent triangles problems by Side-Side-Side