Question

Fig. 6.57
2. In Fig. 6.57, D is a point on hypotenuse AC of A ABC, such that BD LAC, DM I BC
• DN LAB. Prove that :
(i) DM^2= DN.MC
(ii) DN^2 = DM.AN​

Answer 1

Answer:

ANSWER

(i) In △BDC

DM⊥BC

△BMD∼△DMC....(1) [If a perpendicular is drawn the vertex of the right angle of a right triangle to the hypotenuse, the triangle on both sides of the perpendicular are similar to the whole triangle and to each other]

In △BDA

DN⊥AB

△AND∼△DNB....(2) [If a perpendicular is drawn the vertex of the right angle of a right triangle to the hypotenuse, the triangle on both sides of the perpendicular are similar to the whole triangle and to each other]

Using equation (1)

△BMD∼△DMC

BM/DM=MD/MC [If two triangle are similar, then the ratio of their corresponding sides are equal]

DM 2 =BM.MC.....(3)

In △ABC

AB⊥BC and DM⊥BC

⇒AB∣∣DM [Perpendicular drawn to a same line are parallel to each other]

∴NB∣∣DM

Also,

CB⊥CB and DN⊥AB

⇒CB∣∣DN [Perpendicular drawn to a same line are parallel to each other]

∴MB∣∣DN

Now, in quardrilateral DNBM

ND∣∣DM and MB∣∣DN

Since both pairs of opposite sides are parallel

DNMB is a parallelogram

Since opposite sides of parallelogram are equal

∴DN=MB and DM=NB.....(4)

Putting BM=DN in equation (3)

DM 2=BM.MC

DM 2=DN.MC

Hence proved

(ii) From (2)

△AND∼△DNB

AN/DN=DN/BN [If two triangles are similar, then the ratio of their corresponding sides are equal]

DN 2 =AN.BN

Putting BN=DM from (4)

DN 2=AN.DM

Hence proved

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