# Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC).

[Hint :From A and C, draw perpendiculars to BD.]

**Solution:**

Let's draw AM ⊥ BD and CN ⊥ BD. Now we can find the area of triangles to get the required result.

Proof:

ar (ΔABP) = 1/2 × PB × AM........................ (i)

ar (ΔAPD) = 1/2 × PD × AM....................... (ii)

Dividing Equation (ii) by (i), we get,

ar (ΔAPD)/ar (ΔABP) = (1/2 × PD × AM)/(1/2 × PB × AM)

⇒ ar (ΔAPD)/ar (ΔABP) = PD/PB..................................... (iii)

Similarly,

ar (ΔCDP)/ar (ΔBPC) = PD/PB............................. (iv)

From Equation (iii) and (iv), we get

ar (ΔAPD)/ar (ΔABP) = ar (ΔCDP)/ar (ΔBPC)

⇒ ar (ΔAPD) × ar (ΔBPC) = ar (ΔABP) × ar (ΔCDP)

Hence proved.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 9

**Video Solution:**

## Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that; ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC). [Hint :From A and C, draw perpendiculars to BD.]

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.4 Question 6

**Summary:**

If diagonals AC and BD of a quadrilateral ABCD intersect each other at P, then ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC).

**☛ Related Questions:**

- In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
- In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
- In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).[Hint: Join AC.]
- In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show thati) ar (BDE) =1/4 ar (ABC)ii) ar (BDE) = 1/2 ar (BAE)iii) ar (ABC) = 2 ar (BEC)iv) ar (BFE) = ar (AFD)v) ar (BFE) = 2 ar (FED)vi) ar (FED) = 1/8 ar (AFC)[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]