∆abc is isosceles, ab = bc, and ch is an altitude. how long is ac, if ch = 84 cm and m∠hbc = m∠bac +m∠bch?

Given that ∆abc is isosceles and ab = bc, then m<bac = m<acb and m<hbc = 180 - m<bac - m<acb

Given that m∠hbc = m∠bac +m∠bch, then m<hbc + m<bch = m<bac + 2m<bch

But m<hbc + m<bch = 90°, thus 90° = m<bac + 2m<bch

Also, m∠bac + m∠ach = 90° ⇒ m<bac + 2m<bch = m<bac + m<ach ⇒ 2m<bch = m<ach

Since, Δabc is isosceles with ab = bc ⇒ m<bac = m<acb.

Also, m<acb = m<ach + m<bch ⇒ m<acb = 3m<bch = m<bac

Since m<bch : m<ach = 1 : 2 ⇒ bh : ah = 1 : 2

Thus, [tex]bh:bc:ch=1:3:\sqrt{8}[/tex]

Given that ch = 84 cm, then

[tex] \frac{bc}{ch} = \frac{bc}{84} = \frac{3}{\sqrt{8}} \\ \\ \Rightarrow bc= \frac{3\times84}{\sqrt{8}} =63\sqrt{2}=ab[/tex]

Now, [tex]ah= \frac{2}{3} (63\sqrt{2})=42\sqrt{2}[/tex]

[tex]ac= \sqrt{ch^2+ah^2} \\ \\ = \sqrt{84^2+(42\sqrt{2})^2} \\ \\ = \sqrt{7056+3528} = \sqrt{10584} \\ \\ =42\sqrt{6}[/tex]