Question

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Answer 1

area of triangle = 1/2 * b *h

area of triangle ABC = 1/2*BC*AO

similarly

area of triangle DBC = 1/2*BC*OD

area of triangle ABC/area of triangle DBC = AO/DO ( 1/2 & same base BC are cancelled)

(HENCE PROVED)

Answer 2

Step-by-step explanation:

proove :

1. FORMULA OF ∆ABC= 1/2*BC*AL
2. FORMULA OF ∆DCB=1/2* BC*DM

TAKE BOTH DIVISION

$= \frac{ \frac{1}{2} \times bc \times al}{ \frac{1}{2} \times bc \times dm} \\ = \frac{al}{dm}$

IN ∆AOL & ∆DOM

ALO = DMO. ( 90° ANGLE)

AOL = DOM. (Approach ANGLE)

SO,. ∆AOL ~ ∆DOM. (FROM. ANGLE ANGLE condition)

SO,.

AL/DM = AO/DO........(1)

∆ABC/∆DBC = AL/DM

=AO/DO. { FROM (1) }

∆ABC/∆DBC=AO/DO. HENEC

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