Step 1

Using basic properties of triangles we can solve this question.

Step2

angle abc is bisected by bd

angle abg is bisected by be

angle dbe=90(half of a angle in a straight line)

bd=15cm

be=20cm

By pythagoras theorem,

de=25cm

Let de=xcm

\(\displaystyle{b}{e}=\sqrt{{{b}{e}^{{{2}}}-{c}{e}^{{{2}}}}}\)

\(\displaystyle=\sqrt{{{b}{d}^{{{2}}}-{d}{c}^{{{2}}}}}\)

\(\displaystyle={12}{c}{m}\)

Step 3

\(\displaystyle\angle{d}{b}{c}={\frac{{{d}{c}}}{{{b}{c}}}}\)

\(\displaystyle={\frac{{{9}}}{{{12}}}}\)

\(\displaystyle{d}{c}={x}\)

\(\displaystyle{e}{c}={25}-{x}\)

\(\displaystyle{b}{e}^{{{2}}}-{c}{e}^{{{2}}}={b}{d}^{{{2}}}-{x}^{{{2}}}\)

\(\displaystyle{x}={d}{c}\)

\(\displaystyle={9}{c}{m}\)

\(\displaystyle\angle{a}{b}{c}={2}\angle{0}{b}{c}\)

\(\displaystyle={\frac{{{24}}}{{{7}}}}\)

\(\displaystyle{\frac{{{24}}}{{{7}}}}={\frac{{{y}}}{{{12}}}}\)

\(\displaystyle{y}={\frac{{{24}\star{12}}}{{{7}}}}\)

\(\displaystyle{a}{c}={\frac{{{288}}}{{{7}}}}\)

By pythagoras theorem

\(\displaystyle{a}{c}^{{{2}}}+{b}{c}^{{{2}}}={a}{b}^{{{2}}}\)

\(\displaystyle{a}{b}={\frac{{{300}}}{{{7}}}}\)

Using basic properties of triangles we can solve this question.

Step2

angle abc is bisected by bd

angle abg is bisected by be

angle dbe=90(half of a angle in a straight line)

bd=15cm

be=20cm

By pythagoras theorem,

de=25cm

Let de=xcm

\(\displaystyle{b}{e}=\sqrt{{{b}{e}^{{{2}}}-{c}{e}^{{{2}}}}}\)

\(\displaystyle=\sqrt{{{b}{d}^{{{2}}}-{d}{c}^{{{2}}}}}\)

\(\displaystyle={12}{c}{m}\)

Step 3

\(\displaystyle\angle{d}{b}{c}={\frac{{{d}{c}}}{{{b}{c}}}}\)

\(\displaystyle={\frac{{{9}}}{{{12}}}}\)

\(\displaystyle{d}{c}={x}\)

\(\displaystyle{e}{c}={25}-{x}\)

\(\displaystyle{b}{e}^{{{2}}}-{c}{e}^{{{2}}}={b}{d}^{{{2}}}-{x}^{{{2}}}\)

\(\displaystyle{x}={d}{c}\)

\(\displaystyle={9}{c}{m}\)

\(\displaystyle\angle{a}{b}{c}={2}\angle{0}{b}{c}\)

\(\displaystyle={\frac{{{24}}}{{{7}}}}\)

\(\displaystyle{\frac{{{24}}}{{{7}}}}={\frac{{{y}}}{{{12}}}}\)

\(\displaystyle{y}={\frac{{{24}\star{12}}}{{{7}}}}\)

\(\displaystyle{a}{c}={\frac{{{288}}}{{{7}}}}\)

By pythagoras theorem

\(\displaystyle{a}{c}^{{{2}}}+{b}{c}^{{{2}}}={a}{b}^{{{2}}}\)

\(\displaystyle{a}{b}={\frac{{{300}}}{{{7}}}}\)