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two triangles are similar if a line drawn is parallel to one side and intersect two sides in distinct point​

Sagot :

Step-by-step explanation:

Solution

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Given:

DE∣∣BC

To prove that:

AE

EC

=

AD

BD

Proof:

∠AED=∠ACB Corresponding angles

∠ADE=∠ABC Corresponding angles

∠EAD is common to both the triangles

⇒ΔAED∼ΔACB by AAA similarity

AE

AC

=

AD

AB

AE

AE+EC

=

AD

AD+BD

AE

EC

=

AD

BD