Q. 44.6( 7 Votes )

# If in triangle ABC, AD is a median and AM is perpendicular to BC then prove that AB^{2}= AD^{2} – BC × DM + 1/4 BC^{2}.

Answer :

** Given:** D is the mid-point of BC.

And AM ⊥ BC.

__To Prove:__

** Proof:** In right-angled ∆ABM,

By Pythagoras theorem, we can write as

AB^{2} = AM^{2} + BM^{2}

[∵ (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}] …(i)

In right-angled ∆AMD,

By Pythagoras theorem, we can write as

AD^{2} = AM^{2} + MD^{2}

[∵ (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}] …(ii)

Subtract equation (ii) from equation (i), we get

AB^{2} – AD^{2} = AM^{2} – AM^{2} + BM^{2} – MD^{2}

⇒ AB^{2} – AD^{2} = 0 + BM^{2} – MD^{2}

⇒ AB^{2} = AD^{2} – MD^{2} + BM^{2}

⇒ AB^{2} = AD^{2} – DM^{2} + (BD – DM)^{2}

⇒ AB^{2} = AD^{2} – DM^{2} + BD^{2} + DM^{2} – 2(BD)(DM)

⇒ AB^{2} = AD^{2} + BD^{2} – 2(BD)(DM)

[∵ ]

**Hence, proved.**

Rate this question :

Prove that the angles opposite to equal sides of a triangle are equal

RS Aggarwal & V Aggarwal - MathematicsIf the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is

RD Sharma - MathematicsIn Δ *ABC*, if ∠*A* = 100° *AD* bisects ∠*A* and AD⊥BC. Then, ∠*B* =

D is any point on side AC of a ΔABC with AB = AC. Show that CD < BD.

NCERT Mathematics ExemplarIn a triangle *ABC*, if *AB* = *AC* and AB is produced to *D* such that *BD* = *BC*, find ∠*ACD*: ∠*ADC.*

In Δ *ABC*, ∠*A*=50° and *BC* is produced to a point *D*. If the bisectors of ∠*ABC* and ∠*ACD* meet at *E*, then ∠*E* =

Compute the value of *x* in each of the following figures:

(i)

(ii)

(iii)

(iv)

RD Sharma - Mathematics

In Δ *PQR*, if *PQ*=*QR* and *L, M* and *N* are the mid-point of the sides *PQ, QR* and *RP* respectively. Prove that *LN*=*MN*.

In a right-angled triangle, one of the acute measures 53°^{.} Find the measure of each angle of the triangle.

Prove that the perimeter of a triangle is greater than the sum of its altitudes.

RD Sharma - Mathematics