In triangle ABC and DEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Also, AL and DM are medians. Prove that BC/EF = AL/DM.

In Δ ABC and Δ DEF, ⇒ ∠A = ∠D (Given) ⇒ ∠B = ∠E (Given) ⇒ ∠C = ∠F (Given) ∴ Δ ABC ∼ Δ DEF (By AAA postulate) Since, AL and DM are medians of triangles ABC and DEF respectively. ∴ BL = and EM = We know that, Corresponding sides of similar triangles are proportional. In Δ ABL and Δ DEM, ⇒ ∠B = ∠E (Given) ⇒ [From equation (2)] ∴ Δ ABL ∼ Δ DEM (By SAS postulates) As, corresponding sides of similar triangles are proportional. .........(3) From equation (1) and (3), we get : Hence, proved that

In the given figure, the value of x is:
15
30
36
40

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In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.
prove that angle BAC = 90°.

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Mathematics

In triangle ABC and DEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Also, AL and DM are medians. Prove that BC/EF = AL/DM.

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Related Questions

In the given figure, the value of x is:
15
30
36
40

In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.
prove that angle BAC = 90°.

Mathematics

In triangle ABC and DEF, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Also, AL and DM are medians. Prove that BC/EF = AL/DM.

Similarity

Related Questions

In the given figure, the value of x is:15303640

In triangle ABC, AD is perpendicular to side BC and AD2 = BD × CD.prove that angle BAC = 90°.

In the given figure, the value of x is:
15
30
36
40

In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.
prove that angle BAC = 90°.