Question

In the given figure, ABCD is a quadrilateral, AB = 9 cm, BC = 12 cm, CD = 13 cm, ED = 5 cm and AC = 15 cm. Perpendiculars are drawn from B and D to AC, meeting AC at points P and Q respectively. Which of the following is correct?

• A. ΔABC ≅ ΔADC
• B. ΔABC ≅ ΔAEC
• C. ΔAEC ≅ ΔCDE
• D. ΔABE ≅ ΔCQD

Answer 1

The correct option is B ΔABC ≅ ΔAEC


Given: AB = 9 cm, BC = 12 cm, CD = 13 cm, ED = 5 cm and AC = 15 cm
In right-angled triangle CED, by Pythagoras theorem
$(CD{)}^2=(EC{)}^2+(ED{)}^2$
$\Rightarrow (13{)}^2=(EC{)}^2+(5{)}^2$
$\Rightarrow EC=\sqrt{169-25}=\sqrt{144}$
⇒ EC = 12 cm
∴ EC = BC …..(i)
In ΔBPC and ΔEPC,
∠BPC = ∠EPC (Each 90°)
BC = EC [From (i)]
PC = PC (Common)
ΔBPC ≅ ΔEPC (RHS congruence rule) …..(ii)
⇒ BP = PE (CPCT) …..(iii)
In ΔAPB and ΔAPE,
PB = PE [From (iii)]
∠APB = ∠APE (Each 90°)
AP = AP (Common)
∴ ΔAPB ≅ ΔAPE (SAS congruence rule) …..(iv)
From (ii) and (iv), we get
ΔABC ≅ ΔAEC
Hence, the correct answer is option (b).