Advertisements
Advertisements
Question
In ∆ABC, AB = AC. Show that the altitude AD is median also.
Advertisements
Solution
In ∆ABD and ∆ADC,
AB = AC ...(Since is an isosceles triangle)
AD = AD ...(common side)
∠ADB = ∠ADC ...(Since AD is the altitude so each is 90°)
⇒ ΔADB ≅ ΔADC ...(RHS congruence criterion)
BD = DC ...(cpct)
⇒ AD is the median.
APPEARS IN
RELATED QUESTIONS
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
Which congruence criterion do you use in the following?
Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
You want to show that ΔART ≅ ΔPEN,
If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
1) RT = and
2) PN =
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
In Fig. 10.40, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ ΔSAT.
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.
In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.
