Selina solutions

Selina solutions

Grade 7

Congruency: Congruent Triangles | Exercise 19

Question 13

Q13) ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:

(i) AD = BE

(ii) BD = CE

Solution:

In \triangle ABC,

AB = BC = CA

AD\perp BC.\ BE\perp AC

Proof: In \triangle ADC\ and\ \triangle BEC

\angle ADC=\angle BEC\ \ \ \left(each\ 90^{\circ}\right)

\angle ACD=\angle BCE

and AC = BC (sides of an equilateral triangle)

\triangle ADC=\triangle BEC\ \ \ \left(AAS\ Axiom\right)

Hence (i) AD = BE

(ii) BD = CE

Hence proved

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Question 13

Q13) ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:

(i) AD = BE

(ii) BD = CE

Solution:

In \triangle ABC,

AB = BC = CA

AD\perp BC.\ BE\perp AC

Proof: In \triangle ADC\ and\ \triangle BEC

\angle ADC=\angle BEC\ \ \ \left(each\ 90^{\circ}\right)

\angle ACD=\angle BCE

and AC = BC (sides of an equilateral triangle)

\triangle ADC=\triangle BEC\ \ \ \left(AAS\ Axiom\right)

Hence (i) AD = BE

(ii) BD = CE

Hence proved

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