Question

ABCD is a square and P is the midpoint of AD. BP and CP are joined. Prove that PC = PB.


# do not spam

Need correct answer

Answer 1

Answer:

By CPCT

Step-by-step explanation:

1) AB=DC(Sides of a square are equal)

2) AP=PD (P is the mid point of AD)

3) Angle BAP is equal to Angle CDP(all angles of a square are 90 degree)

Therefore BP=CD(By congruent parts of a congruent triangles)

So, we have proved that triangle ABP is congruent to triangle CDP so BP=CP

Please mark me as a brianliest(spelling is wrong)

Answer 2

Your Answer:

Given:-

• ABCD is a square
• P is the mid point of AD

To Prove:-

• PC = PB

Solution:-

$\tt In \ \triangle ABP \ \ and \ \ \triangle DCP$

$\tt \star AB = DC (\because \ AB \ and \ DC \ are \ the \ sides \ of \ square) \\\\ \tt \star AP=DP (\because P \ is \ the \ mid-point \ of \ AD) \\\\ \tt \star \angle BAP = \angle CDP = 90^o [\because \ they \ are \ the \ angles \ of \ \square(square)]$

$\tt So, \ \triangle ABP \ \ and \ \ \triangle DCP \ \ are congruent \ \ with \ \ SAS \ Similarity$

$\tt So, \\\\ \tt BP = PC \ \ \ \ (\because CPCT)$

Hence Proved