Q. 94.3( 148 Votes )

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP

(ii)

Class 10th NCERT - Mathematics 6. Triangles

Answer :

(i)

To Prove: Δ ABC ∼ Δ AMP

Given: In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

Proof:

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ΔABC ΔAMP (By AA similarity)

Hence, Proved.

(ii) Δ ABC ∼ Δ AMP

Now we get that,

Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar. And the converse is also true, so we have

Hence, Proved.

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PREVIOUSE is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB NEXTCD and GH are respectively the bisectors of ∠ ACB and ∠ EGF in such a way that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that: (i) (ii) Δ DCB ~ Δ HGE (iii) Δ DCA ~ Δ HGF

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) Δ ABC ~ Δ AMP (ii)

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP

(ii)