Math, asked by manickamsoundappan, 4 months ago

In a triangle ABC, a line is drawn parallel to BC to meet AB at D and AC at E. DC and BE meet at X. prove that the triangles ADE and ABC are similar.

Answers

Answered by bhellalitadevi
2

Answer:

according to the diagram you described...

according to the diagram you described...∠A=∠A because of the reflexive property

according to the diagram you described...∠A=∠A because of the reflexive propertyCB and ED are parallel

according to the diagram you described...∠A=∠A because of the reflexive propertyCB and ED are parallel∠ACB=∠AED because AE is a transversal and the angles are corresponding angles

according to the diagram you described...∠A=∠A because of the reflexive propertyCB and ED are parallel∠ACB=∠AED because AE is a transversal and the angles are corresponding angles∠ABC=∠ADE because AD is a transversal and the angles are corresponding angles

according to the diagram you described...∠A=∠A because of the reflexive propertyCB and ED are parallel∠ACB=∠AED because AE is a transversal and the angles are corresponding angles∠ABC=∠ADE because AD is a transversal and the angles are corresponding anglesthe three pairs of corresponding angles of the two triangles are congruent (∠A corresponds to itself)

according to the diagram you described...∠A=∠A because of the reflexive propertyCB and ED are parallel∠ACB=∠AED because AE is a transversal and the angles are corresponding angles∠ABC=∠ADE because AD is a transversal and the angles are corresponding anglesthe three pairs of corresponding angles of the two triangles are congruent (∠A corresponds to itself)therefore triangle ABC is similar to triangle ADE .

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