The circle touches AB and AC the lateral sides of the isosceles triangle ABC at the vertices B and C (Fig. 1).
On the arc of this circle, which lies inside this triangle, there is a point K so that the distances from it to the sides AB and AC are equal to 24 cm and 6 cm appropriately (Fig. 2).
Find the distance from point K to side BC.
By using the new geometry, Hyperelliptical Geometry, we will obtain a new value. We must change the current pattern and upgrade appropriate to the physical reality of our Universe.
Construction of a figure with the conditions imposed in the statement: Let's take a point K in the plane. Let two points E1 and E2 be such that KE1=6 and KE2=24. The perpendicular (r) through E1 to KE1 and the perpendicular (s) through E2 to KE2 intersect at A. We construct the two circles (KBC) and (KB'C'), solution of the Apollonian Problem (PRR), which pass through K and are tangent to (r) and a (s). Let F and F' be the orthogonal projections of K onto BC and B'C', respectively. We have KF=KF' and this distance will depend on the location of points E1 and E2. Therefore, the determination of the distance from K to BC, with the conditions imposed, is ambiguous.
Apple GeoGebra: https://amontes.webs.ull.es/geogebra/Liudmyla120524.html
So far, that is, as far as this specific Euclidean geometry case is concerned, let's stick to the old-fashioned stuff involving triangles and angles like:
∢KCF = ∢KBE_2 that are equal because they span the arc BC,
thus △KCF ∼ is similar to ∼△KBE_2, hence,
KC / KB = KF / KE_2, i.e., KC / KB = KF / 24. ((1))
In a similar fashion we get:
∢KBF = ∢KCE_1, and △BKF ∼ △CKE_1, which implies KC/ KB = KE_1 / KF. ((2))
Combining 1-and-2 yields KF/24 = 6/ KF, so KF = 12.
By analogy, in a pure symmetry, we can get KF' = 12.
Sorry, in my previous construction I took the tangents to the circle of diameter KK' from E1, instead of D, to determine the Apollonian circles (KBC) and (KB'C')
Construction of a figure with the conditions imposed in the statement: Let's take a point K in the plane. Let two points E1 and E2 be such that KE1=6 and KE2=24. The perpendicular (r) through E1 to KE1 and the perpendicular (s) through E2 to KE2 intersect at A. We construct the two circles (KBC) and (KB'C'), solution of the Apollonian Problem (PRR), which pass through K and are tangent to (r) and a (s). Let F and F' be the orthogonal projections of K onto BC and B'C', respectively. We have KF=KF' and this distance, 12, does not depend on the location location of points E1 and E2.
Liudmyla Hetmanenko , Steftcho P. Dokov Not necessarly to consider two pairs of similar triangles, because:
quadrilaterals CFKE_1 and BFKE_2 are inscriptible, so
1. angle FKE_1=angle FKE_2 (= 180-C=180-B)
2. angle FE_1K=angle FCK=1/2 arc KB=angle KBE_2=angle KFE_2,
then triangles E_1KF and FKE_2 are similar, and from here KE_1/KF=KF/KE_2 which implies KF = 12
NOTE: Trying to understand Yosef M. Yoely's answer I have some doubts! First, I thought Yosef quickly saw the similarity of triangles E_1KF and FKE_2! If not, maybe Yosef told himself that quadrilaterals CFKE_1 and BFKE_2 are similar because they have equal angles......=>conclusion.....which is not correct, two quadrilaterals with all equal angles are not necessarily similar (for example a square and a rectangle that is not a square !!).