Question

abc is right angle triangle with abc 90 degree, bd perpendicular to ac, dm perpendicular to ab.
Prove:
i) DM^2 = DN x CM
ii) DN^2 = DM x AN

Answer 1

Answer:

i) DM^2 = DN x CM

ii) DN^2 = DM x AN

Step-by-step explanation:

Step 1: A right triangle, also known as a right-angled triangle, right-perpendicular triangle, orthogonal triangle, or previously rectangled triangle, is a triangle with one right angle, or two perpendicular sides. The foundation of trigonometry is the relationship between the sides and various angles of the right triangle.

Step 2: Both an isosceles triangle and a scalene triangle can be right triangles. All three sides of a scalene right triangle will be unequal in length, and any angle will be a right angle. The lengths of the base and perpendicular sides, including the right angle, of an isosceles right triangle will be equal. The hypotenuse will be the third unbalanced side.

We have,

$\mathrm{AB} \perp \mathrm{BC} and \mathrm{DM} \perp \mathrm{BC}$

$\Rightarrow \mathrm{AB} \| \mathrm{DM}$

Similarly, we have

$\mathrm{CB} \perp \mathrm{AB}$ and $\mathrm{DN} \perp \mathrm{AB}$

$\Rightarrow \mathrm{CB} \| \mathrm{DN}$

Hence, quadrilateral BMDN is a rectangle.

$\therefore \mathrm{BM}=\mathrm{ND}$

Step 3: (i) In $\triangle B M D$, we have

$\begin{aligned}& \angle 1+\angle \mathrm{BMD}+\angle 2=180^{\circ} \\& \Rightarrow \angle 1+90^{\circ}+\angle 2=180^{\circ} \\& \Rightarrow \angle 1+\angle 2=90^{\circ}\end{aligned}$

Similarly, in $\triangle D M C$, we have

$\angle 3+\angle 4=90^{\circ}$

Since $B D \perp A C$. Therefore,

$\angle 2+\angle 3=90^{\circ}$

Now, $\angle 1+\angle 2=90^{\circ}$ and $\angle 2+\angle 3=90^{\circ}$

$\begin{aligned}& \Rightarrow \angle 1+\angle 2=\angle 2+\angle 3 \\& \Rightarrow \angle 1=\angle 3\end{aligned}$

Also, $\angle 3+\angle 4=90^{\circ}$and $\angle 2+\angle 3=90^{\circ}$

$\Rightarrow \angle 3+\angle 4=\angle 2+\angle 3 \Rightarrow \angle 2=\angle 4$

Thus, in$\triangle$ 'sBMD and DMC, we have $\angle 1=\angle 3$ and $\angle 2=\angle 4$

So, by AA-criterion of similarity, we have

$\begin{aligned}& \triangle \mathrm{BMD} \sim \triangle \mathrm{DMC} \\\Rightarrow & \frac{\mathrm{BM}}{\mathrm{DM}}=\frac{\mathrm{MD}}{\mathrm{MC}} \\\Rightarrow & \frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{DM}}{\mathrm{MC}} \quad[\because \mathrm{BM}=\mathrm{ND}] \\\Rightarrow & \mathrm{DM}^2=\mathrm{DN} \times \mathrm{MC} \quad[\text { Hence proved }]\end{aligned}$

(ii) Proceeding as in (i), we can prove that

$\begin{aligned}& \triangle \mathrm{BND} \sim \triangle \mathrm{DNA} \\\Rightarrow & \frac{\mathrm{BN}}{\mathrm{DN}}=\frac{\mathrm{ND}}{\mathrm{NA}} \\\Rightarrow & \frac{\mathrm{DM}}{\mathrm{DN}}=\frac{\mathrm{DN}}{\mathrm{AN}} \\\Rightarrow & \left.\mathrm{DN}^2=\mathrm{DM} \times \mathrm{AN} \quad \text { [Hence proved }\right]\end{aligned}$

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