Verified answer

Here's given the correct diagram for the question in the attachment.

Given:

• Two Triangles BAC and BDC drawn on the same base respectively
• ∠A = ∠D = 90°

To Prove:

• AP x PC = DP x PB.

Proof:

∠APB =∠DPC ( Vertically Opposite Angles )

∠BAC =∠BDC ( Vertically Opposite Angles )

so,

( By AA-criterion of similarity )

△APB = △DPC

so,

AP/DP = PB/PC

▪︎AP x PC= DP x PB.

Sure, here is a proof that AP x PC = OP x PB:

• Consider the triangles APB and CPD. Both triangles have a right angle at P and share the same hypotenuse PB. Additionally, since triangles BAC and BDC are right-angled at A and D respectively, and they share the same base BC, we can conclude that angles APB and CPD are congruent.
• Therefore, triangles APB and CPD are similar by the AA similarity criterion. This means that the corresponding sides of the triangles are proportional.

In particular, we have the following proportion:

• AP/OP = PC/PB

Multiplying both sides of this equation by AP and PB, we get:

• AP^2/OP^2 = PC^2/PB^2
• Since AP^2/OP^2 = PC^2/PB^2, we can cross-multiply to get:
• AP x PC = OP x PB
• This proves that AP x PC = OP x PB.

Hope it helps...✨