Question

please solve the question and I will mark your answer as BRAINLIEST.​

Answer 1

Answer:

In a right triangle ABC, right angled at B, D is a point on hypotenuse such that BD⊥AC. If DP⊥AB and DQ⊥BC then. ... since all the angles in quadrilateral BQDP are right angles. Hence, BQDP is a rectangle

so it is proved

Answer 2

Given :-

• ABC is right angled triangle.

• B is 90 ° angle

• D is on hypotenuse

• BD perpendicular to AC

• DP perpendicular to AB

• DQ perpendicular to BC .

Prove that :-

• 1. DQ² = DP × QC

• 2. DP² = DQ × AP .

Solution :-

☆ We know that if if a perpendicular is drawn from the right angled vertex of right angle triangle then the Triangle formed by it are similar to each other as well as similar to the whole Triangle .

Now we'll use the identity in ∆ BDC

Here DQ is drawn drawn from the right angle BDC so from this we conclude that

→ ∆CQD ≈ ∆DQB

Now ifif triangles are similar then the ratio of their corresponding sides will also be similar

$\sf{\implies \frac{CQ}{DQ} = {DQ}{QB} }$

Cross multiple here

$\sf{\implies DQ \times DQ = CQ \times QB \:\:\:\: eq \: 1st}$

Now if we observe the quadrilateral PDQB we note that it's all angles are of 90 degree so it forms a rectangle.

$\sf{\implies DP = QB \: and \: PB = DQ }$

Opposite sides of rectangle are equal .

Now replacing QB by DP in equation first .

${\underline{\underline{\sf{\implies {DQ}^{2} = DP \times CQ }}}}$

Now again using the theorem of perpendicular in ∆ABD .

Here D is 90 degree and DP is drawn from this right angle.

→ ∆APD ≈ ∆DPB

Ratio of their corresponding sides is also equal .

$\sf{\implies \frac{AP}{DP} = \frac{PD}{PB} }$

Cross multiplying

$\sf{\implies {DP}^{2} = AP \times PB }$

We have already prove that PB is equals to DQ , so replacing them

${\underline{\underline{\sf{\implies {DP}^{2} = AP \times DQ }}}}$

Hence both are proved