Question

The side BC of a $△$ABC produced, such that D is on ray BC. The bisector of $\angle$A meets BC in L. Prove that $\angle$ABC + $\angle$ACD = 2$\angle$ALC. [3 MARKS]

Answer 1

Process : 2 Marks
Proof : 1 Mark


In $\Delta ABC$,

$\angle$ACD = $\angle$B + $\angle$A

$\angle$ACD = $\angle$B + 2$\angle$1 ..........(i)

In $\Delta ABL$,

$\angle$ALC = $\angle$B + $\angle$BAL

$\angle$ALC = $\angle$B + $\angle$1

2$\angle$ALC = 2$\angle$B +2$\angle$1 ......(ii)

Subtracting (i) from (ii), we get

2$\angle$ALC - $\angle$ACD = $\angle$B

$\angle$ACD + $\angle$B = 2$\angle$ALC

$\angle$ACD + $\angle$ABC = 2$\angle$ALC