Advertisements
Advertisements
प्रश्न
In figure, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD
Advertisements
उत्तर
Consider the figure Given
AB = AC, DB = DC and given to find the ratio
∠ABD = ∠ACD
Now,∠ABC and ∠DBC are isosceles triangles since AB = AC and
DB = DC respectively
⇒ ∠ABC = ∠ACB and ∠DBC = ∠DCB [ ∵ angles opposite to equal sides are equal]
Now consider,
∠ABD : ∠ACD
⇒ (∠ABC - ∠DBC ) : (∠ACB - ∠DCB)
⇒ (∠ABC - ∠DBC ) : (∠ABC - ∠DBC ) [∵∠ABC - ∠ACB and ∠DBC = ∠DCB]
⇒ 1:1
∴∠ABD :∠ACD = 1:1
APPEARS IN
संबंधित प्रश्न
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
In Figure AB = AC and ∠ACD =105°, find ∠BAC.
In a ΔABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Which of the following statements are true (T) and which are false (F):
If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
In Fig. 10.131, prove that: (i) CD + DA + AB + BC > 2AC (ii) CD + DA + AB > BC
Fill in the blank to make the following statement true.
Difference of any two sides of a triangle is........ than the third side.
Fill in the blank to make the following statement true.
If two sides of a triangle are unequal, then the larger side has .... angle opposite to it.
In the given figure, the sides BC, CA and AB of a Δ ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the Δ ABC.
In the given figure, if EC || AB, ∠ECD = 70° and ∠BDO = 20°, then ∠OBD is
In the given figure, x + y =
In the given figure, if AB ⊥ BC. then x =
The angles of a right angled triangle are
In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to ______.
In ∆PQR, if ∠R > ∠Q, then ______.
In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are ______.
CDE is an equilateral triangle formed on a side CD of a square ABCD (Figure). Show that ∆ADE ≅ ∆BCE.
Find all the angles of an equilateral triangle.
ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Figure). To prove that ∠BAD = ∠CAD, a student proceeded as follows:
In ∆ABD and ∆ACD,
AB = AC (Given)
∠B = ∠C (Because AB = AC)
and ∠ADB = ∠ADC
Therefore, ∆ABD ≅ ∆ACD (AAS)
So, ∠BAD = ∠CAD (CPCT)
What is the defect in the above arguments?
[Hint: Recall how ∠B = ∠C is proved when AB = AC].
