In ABC-ray BD bisects - ABC A-D-C side DE- side BC A-E-B-then ABBC-AECB

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Angle Sum Property for Triangles

In ABC,ray ...

Question

In $△ A B C$ ,ray $B D$ bisects $∠ A B C$ $A - D - C$ side $D E | | s i d e B C$ $A - E - B$ ,then $\frac{A B}{B C} = \frac{A E}{C B}$

True

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False

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Solution

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Similar questions

△

∠ A

B D = \frac{a c}{b + c}

D C = \frac{a b}{b + c}

D

E

C A

C B

△ A B C

C

A E^{2} + B D^{2} = A B^{2} + D E^{2}

A E^{2} + B D^{2} = A B^{2} + D E^{2}

In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If $\frac{A D}{A B}$ = $\frac{A E}{x}$ , Then $x$ is equal to ___.

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