Question

in the given figure abcd is a square p is a point inside it such that PB=PD. prove that CPA is a straight line

Answer 1

Consider ABCD, a square of side 10 cm. P is a point inside ABCD, such tht
PDC is an isosceles triangle with the base angles as 15 deg, so the

Since PC = PD, P must be on the perpendicular bisector of CD. In other words, APB will also be an isosceles triangle.

Area of triangle APD = (PC*PD/2)*sin CPD = [(5.176380902^2)/2]*sin 150 = 6.698729811 sq cm.

Area of triangles APD and BDC = [(5.176380902*10)/2]*sin 75 = 25 sq cm.

The area of PADCBP = 25 + 6.698729811 + 25 = 56.698729811

Now the area of ABP = 10^2 - 56.698729811 = 43.30127019 sq cm.

If ABP is an equilateral triangle its area should be (10*10/2)*sin 60 = 43.30127019 sq cm. And this happens to be so, hence ABP is an equilateral triangle

Answer 2

Answer:

Step-by-step explanation: proof

CDP and BCP

P is common

CB=DC

angle DPC=CPB

so, CDP is congruence to BCP

then, DPA and APB

P is common

AB= DA

DP=PB

DPA=APB

so, DPA is congruence to APB

DPC=BPC

DPA=BPA

DPC+DPA+BPC+BPA=360°

2DPC+2DPA=360°

DPC+DPA=180°

hence, A P C is a straight line