<em>AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)</em>
<em>thus in ΔABC and ΔADC, ∠ABC=∠ADC (given), </em>
<em> ∠BAC=∠CAD [from (1)],</em>
<em>AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC</em>
<em>Hence, by AAS axiom, ΔABC ≅ ΔADC,</em>
<em>Therefore, BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD), since (1)</em>
<em />
Hence, BC=DC proved.
Answer:
11y is the exact value.
Step-by-step explanation:
Answer:
AD = 7
Step-by-step explanation:
Given that the two triangles are similar by the SSS (side side side) postulate, the triangles share the same ratios when it comes to their sides.
We know the values for lines DB, EB and CB, therefore we can solve for AB, and subtract DB to find AD
We can solve the problem by solving for x:

Cross multiply.

Simplify.

Subtract the value of DB to find AD.


Answer:
yes i agree x+78
Step-by-step explanation:
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