Triangle
NCERT Exercise 6.3
Part 2
Question 6: In the given figure, if Δ ABE ≅ Δ ACD, show that Δ ADE ∼ Δ ABC.
Answer: 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;
`(AD)/(DB)=(AE)/(EC)`
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
Answer: 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
Answer: 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
Answer: 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
Answer: 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.
Answer: 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
Answer: 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)`
Answer: Since Δ ABC ∼ Δ AMP;
Hence;
`text(hypotenuse)/text(perpendicular)=h/p`
Or, `(CA)/(BC)=(PA)/(MP)`
Or, `(CA)/(PA)=(BC)/(MP)`
Proved