Advertisements
Advertisements
प्रश्न
In ∆ABC, AB = AC. Show that the altitude AD is median also.
Advertisements
उत्तर
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
संबंधित प्रश्न
In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ΔABC ≅ ΔCDA.
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?
You have to show that ΔAMP ≅ AMQ.
In the following proof, supply the missing reasons.
| Steps | Reasons | ||
| 1 | PM = QM | 1 | ... |
| 2 | ∠PMA = ∠QMA | 2 | ... |
| 3 | AM = AM | 3 | ... |
| 4 | ΔAMP ≅ ΔAMQ | 4 | ... |
In Fig. 10.40, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ ΔSAT.
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
