Question

Please solve this, I have an exam tomorrow and your help will make sure I pass!

Answer 1

Step-by-step explanation:

using rules of congruency:-

in ∆rib and ∆bip

<RBI =<pbi (angles btwn bisectors)

<ipb{90°}=<irb

ib=ib. (common side)

therefor ,∆rib=~∆bip(A-S-A congruency rule)

so,

ip=it(c.p.c.t). ------------(1)

---------------

then in ∆ipc and ∆iqc

<IPC=<iqc (90°)

<PCI=<QCI (angles of bisector)

ic = ic (common side)

therefore,. ∆ipc=~∆iqc (A-S-A congruency rule)

so,. ip= iq (c.p.c t)-----------(2)

using (1) and (2)..

ip = ir = iq. (proved)

(ii) question

in ∆ air and ∆ aiq:-

iq = ir (proved in prev. question)

ai=ai( common side)

<Ari= <aqi (90°)

therefore,

∆air=~∆aiq(S-A-S congruency rule)

so, <IAR=<IAQ (C.P.C.T)

If both angles have same value or bisect the <raq equally. (proved)