NCERT Exercise 6.3

Part 2

Question 6: In the given figure, if Δ ABE ≅ Δ ACD, show that Δ ADE ∼ Δ ABC.


Solution: Since Δ ABE ≅ Δ ACD

Hence, BE = CD

And ∠DBE = ∠ECD

In Δ DBE and Δ ECD

BE = CD (proved earlier)

∠DBE = ∠ECD (proved earlier)

DE = DE (common)

Hence, Δ DBE ≅ Δ ECD

This means; DB = EC

This also means;


Hence; DE || BC

Thus, Δ ADE ∼ Δ ABC proved.

Question 7: In the given figure, altitudes AD and CE of Δ ABC intersect each other at point P. Show that:


(a) Δ AEP ∼ Δ CDP

Solution: In Δ AEP and Δ CDP

∠AEP = ∠CDP (Right angle)

∠APE = ∠CPD (Opposite angles)

Hence; Δ AEP ∼ Δ CDP proved (AAA criterion)

(b) Δ ABD ∼ Δ CBE

Solution: In Δ ABD and Δ CBE

∠ADB = ∠CEB (Right angle)

∠DBA = ∠EBC (Common angle)

Hence; Δ ABD ~ Δ CBE proved (AAA criterion)

(c) Δ AEP ∼ Δ ADB

Solution: In Δ AEP and Δ ADB

∠AEP = ∠ADB (Right angle)

∠EAP = ∠DAB (Common angle)

Hence; Δ AEP ∼ Δ ADB proved (AAA criterion)

(d) Δ PDC ∼ Δ BEC

Solution: In Δ PDC and Δ BEC

∠PDC = ∠BEC (Right angle)

∠PCD = ∠BCE (Common angle)

Hence; Δ PDC ∼ Δ BEC proved (AAA criterion)

Question 8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that Δ ABE ∼ Δ CFB.


Solution: In Δ ABE and Δ CFB

∠ABE = ∠CFB (Alternate angles)

∠AEB = ∠CBF (Alternate angles)

Hence; Δ ABE ∼ Δ CFB proved (AAA criterion)

Question 9: In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:


(a) Δ ABC ∼ Δ AMP

Solution: In Δ ABC and Δ AMP

∠ABC = ∠AMP (Right angle)

∠CAB = ∠PAM (Common angle)

Hence; Δ ABC ∼ Δ AMP proved (AAA criterion)

(a) `(CA)/(PA)=(BC)/(MP)`

Solution: Since Δ ABC ∼ Δ AMP;



Or, `(CA)/(BC)=(PA)/(MP)`

Or, `(CA)/(PA)=(BC)/(MP)`


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