Q.4.
In a right angled AABC, the perpendicular
drawn from the point C on the
hypotenuse meets the hypotenuse at D,
and the bisector of ZC meets the
hypotenuse at E. Prove that AD/DB=AEsquare/EBsquare
Answers
Answer:
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Step-by-step explanation:
We begin with the following lemma:
Lemma: Let XYZ be a triangle with \angle XY Z= 90 + \alpha∠XYZ=90+α. Construct an isosceles triangle XEZ, externally on the side XZ, with base angle.
Then E is the circumcentre of \DeltaΔXYZ.
Proof of the Lemma: Draw ED \perp⊥XZ.
Then DE is the perpendicular bisector of XZ. We also observe that \angle XED = ZED = 90 -\alpha∠XED=ZED=90−α .
Observe that E is on the perpendicular bisector of XZ.
Construct the circumcircle of XY Z. Draw perpendicular bisector of XY and let it meet DE in F.
Then F is thecircumcentre of \DeltaΔXYZ. Join XF.
Then \angle XFD = 90-\alpha∠XFD=90−α.
But we know that \angle XED = 90-\alpha∠XED=90−α. Hence E = F.
Let r_1, r_2r
1
,r
2
and r be the inradii of the triangles ABD, CBD and ABCrespectively. Join PD and DQ. Observe that \angle PDQ = 90^o∠PDQ=90
o
. Hence
PQ^2 = PD^2 + DQ^2 = 2r^2_1 + 2r^2_2PQ
2
=PD
2
+DQ
2
=2r
1
2
+2r
2
2
:
Let s_1s
1
= (AB + BD + DA)=2. Observe that BD = ca=b and AD = \sqrt {AB^2-BD^2}=\sqrt {c^2-\left ( \frac {ca}{b} \right )^2}=c^2/b
AB
2
−BD
2
=
c
2
−(
b
ca
)
2
=c
2
/b.
This gives s_1s
1
= cs=b. But r_1 = s_1r
1
= s
1
c = (c=b)(s b) = cr=b. Similarly, r_2r
2
= ar=b.
Hence PQ^2 = 2r^2\left ( \frac {c^2+a^2}{b^2} \right )=2r^2PQ
2
=2r
2
(
b
2
c
2
+a
2
)=2r
2
Consider \DeltaΔPIQ. Observe thatPIQ = 90+(B=2) = 135PIQ=90+(B=2)=135.
Hence PQsubtends 90^o90
o
on the circumference of the circumcircle of \Delta PIQΔPIQ. But we have seen that \angle PDQ = 90^o∠PDQ=90
o
.
Now construct a circle with PQ asdiameter.
Let it cut AC again in K.It follows that \angle PKQ∠PKQ = 90 and thepoints P, D, K, Q are concyclic.
We also notice \angle KPQ = \angle KDQ = 45^o\,\,and\,\, \angle PQK = \angle PDA = 45^o∠KPQ=∠KDQ=45
o
and∠PQK=∠PDA=45
o
.
Thus PKQ is an isosceles right-angled triangle with KP = KQ.
\therefore KP^2 + KQ^2 = PQ^2 = 2r^2∴KP
2
+KQ
2
=PQ
2
=2r
2
and hence KP = KQ = r.
Now \angle PIQ = 90 + 45\,\, and\,\, \angle PKQ = 2 45^o = 90^o∠PIQ=90+45and∠PKQ=245
o
=90
o
with KP = KQ = r.
Hence K is the circumcentre of \Delta PIQΔPIQ.
(Incidentally,
This also shows that KI = r and hence K is the point of contact of the incircle of \DeltaΔABC with AC.)
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