Question

4. Measures of some angles in the figure
are given. Prove that
AP/ AQ
PB /QC........ plz answer​

Answer 1

Answer:

∠P = 60° and ∠B = 60°.

To prove:

AP/PB = AQ/QC

Proof:

1) In the given triangle ABC

ΔABC & ΔAPQ

∠P = ∠B = 60° (given)

∠A = ∠A ( common)

∠C = ∠Q (ultimately equal if the other two are equal)

ΔABC ≈ ΔAPQ ( by AA similarity rule)

2) So by the similarity ration we get

\frac{AB}{AP } = \frac{AC}{AQ} =\frac{BC}{PQ}

AP

AB

=

AQ

AC

=

PQ

BC

\frac{AB}{AP} -1=\frac{AC}{AQ}-1

AP

AB

−1=

AQ

AC

−1 (Subtract 1 from both side)

\frac{AB-AP}{AP}=\frac{AC-AQ}{AQ}

AP

AB−AP

=

AQ

AC−AQ

\frac{PB}{AP}=\frac{QC}{AQ}

AP

PB

=

AQ

QC

\frac{AP}{PB}=\frac{AQ}{QC}

PB

AP

=

QC

AQ

(reciprocal)

Hence proved.

Answer 2

Answer:

follow me

Step-by-step explanation:

then I will answer